If the sum of the first 2n terms of 2, 5, 8, | vedclass

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(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is

Vedclass · Accepted Answer

$11$

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(C) For the first arithmetic progression $2, 5, 8, \dots$,the first term $a_1 = 2$ and common difference $d_1 = 3$. The sum of the first $2n$ terms is $S_{2n} = \frac{2n}{2} [2(2) + (2n - 1)3] = n[4 + 6n - 3] = n(6n + 1)$.For the second arithmetic progression $57, 59, 61, \dots$,the first term $a_2 = 57$ and common difference $d_2 = 2$. The sum of the first $n$ terms is $S_n = \frac{n}{2} [2(57) + (n - 1)2] = \frac{n}{2} [114 + 2n - 2] = \frac{n}{2} [112 + 2n] = n(56 + n)$.Given that $S_{2n} = S_n$,we have $n(6n + 1) = n(56 + n)$.Since $n 
eq 0$,we divide by $n$: $6n + 1 = 56 + n$.$5n = 55$,which gives $n = 11$.

If the sum of the first $2n$ terms of $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of $57, 59, 61, \dots$,then $n$ is equal to

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